How do you find the asymptotes for #f(x)= x/(x(x-2))#?

1 Answer
Mar 28, 2016


Vertical asymptotes, horizontal asymptotes and "holes" can be found in this function.


First, cancel the factor of x from both the numerator and denominator of the expression:
#f(x) = 1/(x-2)#

That factor being removed, causes a "hole" in the graph. This is sometimes called a removable discontinuity. So, at #x!=0#, there is a "hole".

To find a vertical asymptote, set the denominator equal to 0 and solve:
#x - 2 = 0# so x = 2 is the vertical asymptote.

To find a horizontal asymptote, notice that the degree of x in the denominator is higher than the degree of x in the numerator. That is, #x^0# in the numerator and #x^1# in the denominator. Therefore, large positive or negative values of x will cause the expression to approach 0. That means that a horizontal asymptote will be y = 0.

Still not sure about this? Substitute in large positive and negative values for x like 1000000 or -1000000...Your output values will be very, very small!